Integral Solutions of Non- Homogeneous Quintic Equation with Three Unknowns

ISSN: 2319-8753

International Journal of Innovative Research in Science, Engineering and Technology Vol. 2, Issue 4, April 2013

INTEGRAL SOLUTIONS OF NON- HOMOGENEOUS QUINTIC EQUATION WITH THREE UNKNOWNS

x 2  y 2  xy  x  y 1  (k 2  3)n z 5 M.A.Gopalan 1 , G.Sumathi 2 , S.Vidhyalakshmi 3 Professor, PG and Research, Department of Mathematics, Shrimati Indira Gandhi, College,Trichy-620002, 1 Tamilnadu,India Lecturer, PG and Research, Department of Mathematics, Shrimati Indira Gandhi College , Trichy-620002, Tamilnadu,India 2 Professor, PG and Research, Department of Mathematics , Shrimati Indira Gandhi College, Trichy-620002, 3 Tamilnadu,India

ABSTRACT: The non-homogeneous quintic equation with three unknowns represented by the diophantine equation 2 2 2 n 5 x  y  xy  x  y 1 (k  3) z is analyzed for its patterns of non-zero distinct integral solutions and three different methods of integral solutions are illustrated. Various interesting relations between the solutions and special numbers, namely, polygonal numbers, Jacobsthal numbers, Jacobsthal-Lucas number,Pronic numbers, Stella octangular numbers, Octahedral numbers, Centered Polygonal numbers, Centered Pentagonal Pyramidal numbers, Centered Hexagonal Pyramidal numbers, Generalized Fibonacci and Lucas sequences, Fourth Dimensional Figurate numbers and Fifth Dimensional Figurate numbers are exhibited.

KEYWORDS: Integral solutions, Generalized Fibonacci and Lucas sequences, Quintic non-homogeneous equation with three unknowns M.Sc 2000 mathematics subject classification: 11D25 NOTATIONS

tm,n : Polygonal number of rank n with size m

Prn : Pronic number of rank

Son : Stella octangular number of rank n

jn : Jacobsthal lucas number of rank

J n : Jacobsthal number of rank

GFn : Generalized Fibonacci sequence number of rank

GL n : Generalized Lucas sequence number of rank

Ctm,n : Centered Polygonal number of rank n with size m

ISSN: 2319-8753

International Journal of Innovative Research in Science, Engineering and Technology Vol. 2, Issue 4, April 2013

Cf3,n,5 : Centered Pentagonal Pyramidal number of rank n

Cf3,n,6 : Centered Hexogonal Pyramidal number of rank n

f4,n,3 : Fourth Dimensional Figurate Traingular number of rank n

f4,n,4 : Fourth Dimensional Figurate Square number of rank n

f4,n,6 : Fourth Dimensional Figurate Hexogonal number of rank n

f5,n,3 : Fifth Dimensional Figurate Traingular number of rank n

f5,n,7 : Fifth Dimensional Figurate Heptagonal number of rank n

I.INTRODUCTION The theory of diophantine equations offers a rich variety of fascinating problems. In particular,quintic equations, homogeneous and non-homogeneous have aroused the interest of numerous mathematicians since antiquity[1-3].For illustration, one may refer [4-5] for quintic equation with three unknowns and [6-8] for quintic equation with five unknowns. This communication concerns with an interesting non-homogeneous ternary quintic equation with three unknowns represented by 2 2 2 n 5 x  y  xy  x  y 1 (k  3) z for determining its infinitely many non-zero integral points. Three different methods are illustrated. In method 1, the solutions are obtained through the method of factorization. In method 2, the binomial expansion is introduced to obtain the integral solutions. In method 3, the integral solutions are expressed in terms of Generalized Fibonacci and Lucas sequences along with a few properties in terms of the above integer sequences Also, a few interesting relations among the solutions are presented.

II.METHOD OF ANALYSIS

The Diophantine equation representing a non-homogeneous quintic equation with three unknowns is

(1) Introducing the linear transformations x  u  v, y  u v (2) in (1), it leads to 2 2 2 n 5 (u 1)  3v  (k  3) z (3) The above equation (3) is solved through three different methods and thus, one obtains three distinct sets of solutions to (1).

A. Method:1 2 2 Let z  a  3b (4)

Substituting (4) in (3) and using the method of factorization,define n 5 (u 1)  i 3v (k  i 3) (a  i 3b) (5)

ISSN: 2319-8753

International Journal of Innovative Research in Science, Engineering and Technology Vol. 2, Issue 4, April 2013

n 5  r exp(in)(a  i 3b)

2 1 3 where r  k  3,   tan (6) k Equating real and imaginary parts in (5) we get n 2 5 3 2 4 4 2 3 5 u  (k  3) 2 cosn(a  30a b  45ab )  3 sin(5a b  30a b  9b ) n 2  4 2 3 5 1 5 3 2 4  v  (k  3) 2 cosn (5a b  30 a b  9b )  sin (a  30 a b  45ab )  3  Substituting the values of u and v in (2), the corresponding values of x,, y z are represented by

 n cosn(a5  30a3b2  45ab4  5a4b  30a2b3  9b5)   2   2 x(a,b,k)  (k  3) sin 5 3 2 4 4 2 3 5  1   (a  30a b  45ab 15a b  90a b  27b )    3 

 n cosn(a5  30a3b2  45ab4  5a4b  30a2b3  9b5 )   2   2 y(a,b,k)  (k  3) sin 5 3 2 4 4 2 3 5  1   (a  30a b  45ab 15a b  90a b  27b )    3  z(a,b)  a 2  3b 2 A few numerical examples are given below: Table: Numerical Examples: n k a,b x y z 0 1 1,1 -1 31 4 1 2 1,1 63 95 4 2 2 1,1 255 159 4 3 2 2,1 1264 2768 7

B. Method 2: n Using the binomial expansion of (k  i 3) in (5) and equating real and imaginary parts, we have 5 3 2 4 4 2 3 5 u 1  f ()(a 30a b  45ab ) 3g()(5a b 30a b  9b ) 4 2 3 5 5 3 2 4 v  f ()(5a b 30a b  9b )  g()(a 30a b  45ab )

Where

ISSN: 2319-8753

International Journal of Innovative Research in Science, Engineering and Technology Vol. 2, Issue 4, April 2013

n    2  f ()  (1)r n k n2r (3)r   C2r  r 0  ...... (7) n1     2   g()  (1)r 1n k n2r 1(3)r 1  C2r1 r 1 

In view of (2) and (7) the corresponding integer solution to (1) is obtained as

5 3 2 4 4 2 3 5 x   f ()  g()(a 30a b  45ab )   f () 3g()(5a b 30a b  9b )1

5 3 2 4 4 2 3 5 y   f ()  g()(a 30a b  45ab )   f ()  3g()(5a b 30a b  9b )1 2 2 z  a  3b

C. Method 3:

Taking n  0 and u 1U in (3), we have, 2 2 5 U  3v  z (8) Substituting (4) in (8), we get 2 2 2 2 5 U  3v  (a  3b ) (9) whose solution is given by 5 3 2 4 U0  (a  30a b  45ab ) 4 2 3 5 v0  (5a b  30a b  9b ) Again taking n 1 in (3), we have 2 2 2 2 2 5 U  3v  (k  3)(a  3b ) (10) whose solution is represented by U1  ku0  3v0

v1  u0  kv0 The general form of integral solutions to (1) is given by  A 3B   s  s  U s   2 2i U0      , s  1,2,3.....  vs   Bs As  v0     2i 3 2  s s Where As  (k  i 3)  (k  i 3) s s Bs  (k  i 3)  (k  i 3)

ISSN: 2319-8753

International Journal of Innovative Research in Science, Engineering and Technology Vol. 2, Issue 4, April 2013

Thus, in view of (2), the following of integers interms of Generalized Lucas and fiboanacci sequence satisfy (1) xs , ys are as follows: 1  GL (2k,k 2  3)(a5  30a3b2  45ab4  5a4b  30a2b3  9b5 )  2 n  xs    1 2 5 3 2 4 4 2 3 5 GFn (2k,k  3)(a  30a b  45ab 15a b  90a b  27b ) 

1 2 5 3 2 4 4 2 3 5  GLn (2k,k  3)(a  30a b  45ab  5a b  30a b  9b )  2  ys    1 2 5 3 2 4 4 2 3 5 GFn (2k,k  3)(a  30a b  45ab 15a b  90a b  27b ) 

The above values of xs , ys satisfy the following recurrence relations respectively 2 xs2  2kxs1  (k  3)xs  0

2 ys2  2kys1  (k  3)ys  0 Properties 2 2 1. xs (a,b,k)  ys (a,b,k)  GLn (2k,k  3)u0  6GFn (2k,k  3)v0  2  2 2 2  2kGLn (2k,k  3) (k  3)GLn1(2k,k  3)u0  2.x (a,b,k)  y (a,b,k)     2 s1 s1 2 2 2 (12k)GFn (2k,k  3)  6(k  3)GFn1(2k,k  3)vo   2 2 3 2  (3k 1)GLn (2k,k  3) (2k  6k)GLn1(2k,k  3)u0  3.x (a,b,k)  y (a,b,k)     2 s2 s2 2 2 3 2 (18k 1)GFn (2k,k  3)  (12k  36k)GFn1(2k,k  3)vo 

83i 3B 4.y (a,a,k)  x (a,a,k)  s 24 f 144 f  25Cf  2Ct  t   16 A (t )(Cf ) s s 6 5,n,7 4,n,3 3,n,6 36,a 4,a s 4,a 3,n,6

i3 3Bs 5.x (a,a,k)  y (a,a,k)  2  (24 f 144 f  30Cf  80t  6t  5t ) s s 2 5,n,7 4,n,3 3,n,5 3,a 7,a 4,a As  (24 f 144 f  25Cf  36Pr  24t ) 2 5,n,7 4,n,3 3,n,6 a 4,a

4  83i 3Bs  6.2As ys (a,a,k)  xs (a,a,k)  24 f5,n,7 144 f4,n,3  25Cf3,n,6  2Ct36,a  t4,a  is a  6  quintic integer.

ISSN: 2319-8753

International Journal of Innovative Research in Science, Engineering and Technology Vol. 2, Issue 4, April 2013

7.Each of the following is a nasty number. 2 2 (a) 6z(a ,a )  i3 3Bs  x (a,a,k)  y (a,a,k)  2  (24 f 144 f  30Cf  80t  6t  5t )  s s 2 5,n,7 4,n,3 3,n,5 3,a 7,a 4,a (b)2As    A  s (24 f 144 f  25Cf  36Pr )  2 5,n,7 4,n,3 3,n,5 a 

8. xs (a,1,k)  ys (a,1,k)  2  As (24 f5,n,7  72 f4,n,4  20Cf3,n,6  Ct104,a  33t4,a ) 

i 3Bs (24 f4,n,7  70Soa 12Ct4,a  29Pra 32t4,a )  A  s (48 f 18Ct  50Pr  78t  28Cf )   2 4,n,7 6,a a 4,a 3,n,6  9.xs (a,1,k)  ys (a,1,k)    i 3  (120 f4,n,6  240 f5,n,3  43Ct4,a  70Cf3,n,6 128Pra 102t4,a )  6  a a 10. z(2 ,2 ) j2a  9J2a  2

III. CONCLUSION To conclude, one may search for other pattern of solutions and their corresponding properties.

REFERENCES 1. L.E.Dickson, History of Theory of Numbers, Chelsea Publishing company, Vol.11, New York (1952). 2 .L.J.Mordell, Diophantine equations, Academic Press, London(1969).

3. Carmichael ,R.D.,The theory of numbers and Diophantine Analysis,Dover Publications, New York (1959) 3 3 5 4 .M.A.Gopalan & A.Vijayashankar, An Interesting Dio.problem x y2 z , Advances in Mathematics, Scientific Developments and Engineering Application, Narosa Publishing House, , 2010,Pp 1-6. 2 2 5 5 .M.A.Gopalan & A.Vijayashankar, Integrated solutions of ternary quintic Diophantine equation x(2 k  1) y  z ,International Journal of Mathematical Sciences 19(1-2),(jan-june 2010), 165-169. 5 6. M.A.Gopalan & A.Vijayashankar, Integrated solutions of non-homogeneous quintic equation with five unknowns xy zw R , Bessel J.Math.,1(1), 2011,23-30. 4 4 2 2 3 7. M.A.Gopalan & A.Vijayashankar, solutions of quintic equation with five unknowns x y 2( z  w ) P ,Accepted for Publication in International Review of Pure and Applied Mathematics. 8. M.A.Gopalan G.Sumathi and S.Vidhyalakshmi, On the non-homogeneous quintic equation with five 3 3 3 3 5 unknowns x  y  z  w  6T ,Accepted for Publication in rijmie journal.